In 8s, a simple 5-link oddagon(*) having three guardians(#).Guardians 8r1c6,r3c8 directly see 8r3c5, while guardian 8r6c6 needs a network to remotely see 8r3c5. But, guardian 8r6c6 does directly see 8r9c6. Since at least one guardian must be true, let's write the following kraken column:

We've avoided the long network solution, but is the logic valid? My old brain wants to say yes, but SpaCe pointed out for a similar situation here that this is probably not the case. There does exist a derived weak inference between the two stte digits, so at least one must be false, but we have not shown a "common outcome" that would definitively eliminate one of the two digits. Do the Forum rules for stte eliminations specifically require that common outcome? Comments would be most welcome!

Sudtyro2 wrote:My old brain wants to say yes, but SpaCe pointed out for a similar situation here that this is probably not the case. There does exist a derived weak inference between the two stte digits, so at least one must be false, but we have not shown a "common outcome" that would definitely eliminate one of the two digits. Do the Forum rules for stte eliminations specifically require that common outcome? Comments would be most welcome!

Hi Steve,I noticed SpAce's doubt at the time and thought I had to dig it. Your question is a good opportunity. The situation is different between your link and the above.

First a side remark: what you need between the two ste digits isn't a weak link "digit1 - digit2" but an inference link "digit1 => digit2", such that if digit1 is True, set1 of guardians is false, digit2 is true, set2 of guardians is false; all guardians are false (contradiction). On September 18 puzzle, such an inference link was available trough a kite. But in the above puzzle, it is not immediate (there is one through a kraken, as you mention). My suggestion of short-writing on September 18 was made with the hypothesis of implicit existence of the link.

So SpAce was right, the logic needs to explicit some inference link between "the two stte digits". The common outcome must be: "anyhow, all guardians are false".

Sudtyro2 wrote:In 8s, a simple 5-link oddagon(*) having three guardians(#).Guardians 8r1c6,r3c8 directly see 8r3c5, while guardian 8r6c6 needs a network to remotely see 8r3c5. But, guardian 8r6c6 does directly see 8r9c6. Since at least one guardian must be true, let's write the following kraken column:

We've avoided the long network solution, but is the logic valid? My old brain wants to say yes, but SpaCe pointed out for a similar situation here that this is probably not the case. There does exist a derived weak inference between the two stte digits, so at least one must be false, but we have not shown a "common outcome" that would definitely eliminate one of the two digits. Do the Forum rules for stte eliminations specifically require that common outcome? Comments would be most welcome!

I would welcome others' comments too, because I already said my piece earlier. I haven't changed my mind, either. In my mind there are two valid ways to achieve a definite conclusion with any kraken-like logic (by "kraken" I mean any SIS, such as the set of oddagon guardians): 1) either all members of the SIS reach a consensus about some conclusion, or 2) some do and the rest self-destruct with a contradiction (but that variant doesn't seem as elegant). In both of those cases we're left with a single conclusion that must be true and we can confidently take that. If it also happens to be an stte-solution, then great, we're done.

However, if we reach two or more different conclusions (none of which is a contradiction), as in your case, it doesn't prove anything directly useful. All you know is that at least one of those conclusions must be true but that's not any different from our starting point (at least one of the SIS members must be true, but we don't know which one(s) -- that's why we try to find a point where they all agree). As I said before, it shouldn't play any part if you happen to know that all of those conclusions are valid stte-eliminations, because that's not something that the proof tells you or what a purely manual solver would be able to know. You should actually run (and notate) your chains all the way to the end to prove that they all do lead to the solution, but that wouldn't be a very satisfactory proof. On the other hand, if you use that prior knowledge without proving it, then it's a case of circular reasoning, as I said before.

Therefore, I still think that we can't accept as a solution two or more different stte-conclusions that have an OR relationship, no matter how elegant it might look, because it depends on knowing beforehand that they in fact are all stte-conclusions. That's not something a manual solver would know without running trials for all paths to the end.

Hi Steve! Related to the recent matrix discussion here I did some practicing with this interesting puzzle and the various solutions (partly because Cenoman had so nicely provided matrices to compare with). I like yours very much, especially since it's the most difficult one to turn into a matrix, but I can't quite agree with the last combo node of the chain. I understand that it's just taking a shortcut, but being a nitpicker, I wanted to try some options:

Needless to say I like the last one best (and the first multi-header the least). What do you think? (It can also be combined nicely with eleven's solution, but that's behind the above link.)

Cenoman, I ran into the same problems as you when trying to convert that into a matrix. I guess your double-matrix (is it a case of BTM?) is the cleanest way to do it. The only alternative I can think of is something like this:

(Don't ask me to explain the rank regions -- it has so many triplets that I get confused.)

Would that work? The difficulty seems to arise from the fact that we have two link triplets (8r3c5 and 8r6c5) working in opposite directions (pointing at each other). My solution is to represent one of them as an OR-node. It's more obvious if we look at it as a net:

Only minor comments:First, an opinion on a point I am not asked to comment: from the four options tried by SpAce, the one I like best is #3. (my reason: it is the closest of the net and the matrix)Second,

If I counted right it has an impressive rank of 4, which makes it a pretty interesting specimen. Maybe one of these days I'll figure out (or someone tells me) how those triplet rules work here (I'm just assuming the eliminations must be Rank 1). Interestingly, 8r3c5 is being linked in three different ways (r3,c5,b2).

I have no problem with that. It does make it more obvious that the two cells at the end aren't connected and where the digits lie. The same information can be inferred from my style but it's not as explicit.

SpAce wrote:I counted the first two lines as one : 3r5c4&5r7c5 8r5c4 3r7c5.I guess that still wouldn't make it a BTM, though its exact definition still eludes me.

Oups ! In my focusing on native strong links, I forgot to explore that one !Edit: after reading SpAce further comments, the matrix cannot be identified as a BTMApplying the process that you have explored in the other thread (here), I guess this matrix can be identified as a BTMI first thought this matrix could be identified as a BTM:

Of course, I am not sure whether the first term in the first row can be split that way. It seems to make sense with the subsequent OR of the two matrices, which maintains an "OR" operator between 8r5c4 and 3r7c5 and thus requires the AND operator to link 3r5c4 and 5r7c5.I now think the first term in the first row cannot be split that way. It does not make sense with a subsequent OR of the two matrices, as there is no native (or derived) "OR" operator between 8r5c4 and 3r7c5. The AND operator to link 3r5c4 and 5r7c5 is required to demonstrate both eliminations -3r9c4 & -5r9c4 and would result of such an OR between 8r5c4 and 3r7c5, which unfortunately is not present.